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수열의 귀납적 정의

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Sundstrom, Mathematical Reasoning · 4.3절 · 2a번영어 원문

Assume that f1,f2,…,fn,…f_1 ,f_2 , \ldots ,f_n , \ldots are the Fibonacci numbers. Prove each of the following:

For each n∈Nn \in \mathbb{N}, f4nf_{4n} is a multiple of 3.

정답·풀이 보기 (영어 원문)

Let P(n)P( n ) be, “f4nf_{4n} is a multiple of 3.” Since f4=3f_4 = 3, P(1)P( 1 ) is true. If P(k)P( k ) is true, then there exists an integer mm such that f4k=3mf_{4k} = 3m. We now need to prove that P(k+1)P(k+1) is true or that f4(k+1)f_{4(k+1)} is a multiple of 3. We use the following:

f4(k+1)=f4k+4=f4k+3+f4k+2=(f4k+2+f4k+1)+(f4k+1+f4k)=f4k+2+2f4k+1+f4k=(f4k+1+f4k)+2f4k+1+f4k=3f4k+1+2f4k\begin{aligned}f_{4 \left( k + 1 \right)} & = f_{4k + 4} \\ & = f_{4k+3} + f_{4k+2} \\ & = \left( f_{4k+2} + f_{4k+1} \right) + \left( f_{4k+1} + f_{4k} \right) \\ & = f_{4k+2} + 2 f_{4k+1} + f_{4k} \\ & = \left( f_{4k+1} + f_{4k} \right) + 2 f_{4k+1} + f_{4k} \\ & = 3f_{4k + 1} + 2f_{4k}\end{aligned}

We now use the assumption that f4k=3mf_{4k} = 3m and the last equation to obtain f4(k+1)=3f4k+1+2⋅3mf_{4(k+1)} = 3f_{4k + 1} + 2\cdot 3m and hence, f4(k+1)=3(f4k+1+2m)f_{4(k+1)} = 3 \left( f_{4k +1} + 2m \right). Therefore, f4(k+1)f_{4(k+1)} is a multiple of 3 and this completes the proof of the inductive step.

한 번 더 — 비슷한 문제Sundstrom, Mathematical Reasoning · 4.3절 · 2c번영어 원문

Assume that f1,f2,…,fn,…f_1 ,f_2 , \ldots ,f_n , \ldots are the Fibonacci numbers. Prove each of the following:

For each n∈Nn \in \mathbb{N} with n≥2n \geq 2, f1+f2+⋯+fn−1=fn+1−1f_1 + f_2 + \cdots + f_{n - 1} = f_{n + 1} - 1.

다 풀었으면 정답 확인

Let P(n)P( n ) be, “f1+f2+⋯+fn−1=fn+1−1f_1 + f_2 + \cdots + f_{n - 1} = f_{n + 1} - 1.” Since f1=f3−1f_1 = f_3 - 1, P(2)P ( 2 ) is true. For k≥2k \geq 2, if P(k)P ( k ) is true, then f1+f2+⋯+fk−1=fk+1−1f_1 + f_2 + \cdots + f_{k - 1} = f_{k + 1} - 1. Then

(f1+f2+⋯+fk−1)+fk=(fk+1−1)+fk=(fk+1+fk)−1=fk+2−1\begin{aligned}\left( f_1 + f_2 + \cdots + f_{k - 1} \right) + f_k & = \left( f_{k + 1} - 1 \right) + f_k \\ & = \left( f_{k+1} + f_k \right) - 1 \\ & = f_{k+2} - 1\end{aligned}

. This proves that if P(k)P \left( k \right) is true, then P(k+1)P \left( k + 1\right) is true.

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